Apply qualitatively and quantitatively Newton's Law of Universal Gravitation, F = G m_1 m_2 / r^2, to determine the magnitude of force, gravitational field strength g = G M / r^2, and acceleration due to gravity at different points in a radial gravitational field

Gravitational field strength at distance $r$ from the centre of a mass $M$:

$g = \frac{G M}{r^2}$.

The distance from the centre is $r = R_E + h = 6.37 \times 10^6 + 1.0 \times 10^6 = 7.37 \times 10^6$ m.

$g = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(7.37 \times 10^6)^2}$
$g = \frac{3.98 \times 10^{14}}{5.43 \times 10^{13}}$
$g = 7.33 \text{ m/s}^2$.

Markers reward the correct use of $r = R_E + h$ (not just $h$), the substitution shown explicitly, and the final answer with units. A common student error is using altitude alone as $r$.

The gravitational force follows the inverse-square law $F = \frac{G M m}{r^2}$. The astronaut is $400$ km above the surface, so the distance from Earth's centre changes from $6370$ km to $6770$ km. The ratio of forces is:

$\frac{F_{\text{ISS}}}{F_{\text{surface}}} = \left(\frac{6370}{6770}\right)^2 \approx 0.89$.

So gravity at the ISS is still about 89% of its surface value, around $8.7 \text{ m/s}^2$.

The astronaut feels weightless not because gravity is absent, but because both the astronaut and the station are in free fall around the Earth. They share the same gravitational acceleration, so there is no normal force between the astronaut and the floor, and no sensation of weight. Apparent weightlessness is a consequence of free fall, not the absence of gravity.

Markers reward the quantitative comparison, the distinction between gravitational force and apparent weight, and the link to free fall.

Word statement. Every pair of point masses attracts each other with a force directed along the line joining them, proportional to the product of the masses and inversely proportional to the square of the distance between their centres.

Equation. $F = \dfrac{Gm_1m_2}{r^2}$, where $F$ is the gravitational force in newtons, $m_1$ and $m_2$ are the two masses in kilograms, $r$ is the distance between their centres in metres, and $G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ is the universal gravitational constant.

Marks: one for a correct word statement (inverse-square, proportional to both masses), one for the correct equation with every symbol defined and the value of $G$ given.

$F = \dfrac{Gm_1m_2}{r^2} = \dfrac{(6.67 \times 10^{-11})(1000)(1000)}{(2.0)^2}$

$F = \dfrac{6.67 \times 10^{-5}}{4.0} = 1.7 \times 10^{-5}\ \text{N}$.

Marks: one for the correct formula with values substituted, one for correctly squaring $r$, one for the answer stated to two significant figures with the unit newton.

$g = \dfrac{GM}{r^2} = \dfrac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}$

$g = \dfrac{4.90 \times 10^{12}}{3.03 \times 10^{12}} = 1.6\ \text{m s}^{-2}$.

Marks: one for the formula $g = GM/r^2$, one for substituting the Moon's own mass and radius (not Earth's), one for the answer $1.6\ \text{m s}^{-2}$ with the unit. Note this is about one-sixth of Earth's surface value.

(a) $r = R_E + h = 6.37 \times 10^6 + 3.0 \times 10^6 = 9.37 \times 10^6\ \text{m}$.

(b) $F = \dfrac{GM_Em}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)}{(9.37 \times 10^6)^2}$

$F = \dfrac{1.99 \times 10^{17}}{8.78 \times 10^{13}} = 2.3 \times 10^3\ \text{N}$.

Marks: one for correctly finding $r = R_E + h$, one for the force formula with values substituted, one for correctly squaring $r$, one for the answer $2.3 \times 10^3\ \text{N}$ with the unit.

(a) The curve falls steeply at first and then flattens out, approaching (but never reaching) zero. This is the inverse-square relationship $g = GM/r^2$: $g$ is proportional to $1/r^2$, so it decreases rapidly with distance but never becomes exactly zero.

(b) The marked data point at $r = 2R_E$ reads $g \approx 2.5\ \text{m s}^{-2}$.

(c) $r = 2R_E = 2(6.37 \times 10^6) = 1.274 \times 10^7\ \text{m}$.

$g = \dfrac{GM_E}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(1.274 \times 10^7)^2} = \dfrac{3.98 \times 10^{14}}{1.62 \times 10^{14}} = 2.5\ \text{m s}^{-2}$.

Marks: one for describing the steep-then-flattening inverse-square shape, one for correctly reading $g \approx 2.5\ \text{m s}^{-2}$ off the graph, one for the substitution $r = 2R_E = 1.274 \times 10^7\ \text{m}$, one for the calculated value matching the graph reading.

(a) $F = \dfrac{GM_Em}{R_E^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(70)}{(6.37 \times 10^6)^2} = \dfrac{2.79 \times 10^{16}}{4.06 \times 10^{13}} = 6.9 \times 10^2\ \text{N}$.

(b) $W = mg = 70 \times 9.8 = 6.9 \times 10^2\ \text{N}$, matching part (a). This confirms that $g = GM_E/R_E^2$ is simply the constant of proportionality between an object's mass and its weight at the surface.

Marks: one for the correct force calculation with substitution, one for the answer $6.9 \times 10^2\ \text{N}$ with the unit, one for showing $mg$ gives the same value and stating the link between the two formulas.

Band-6 plan. (1) State the law and its inverse-square, universal character. (2) Apply it to the apple: near Earth's surface, $r \approx R_E$, giving $g \approx 9.8\ \text{m s}^{-2}$, a straight-line fall. (3) Apply it to the Moon: same law, much larger $r$, giving a much smaller acceleration that curves the Moon's straight-line motion into an orbit rather than a fall to the surface. (4) State the unifying insight: the SAME force law, at different distances, explains both a fall and an orbit. Use numbers for both cases.

Model answer. Newton's Law of Universal Gravitation states that any two masses attract each other with a force $F = Gm_1m_2/r^2$, directed along the line joining them and dependent only on the masses and the separation - the same law applies to every pair of masses in the universe, from an apple to a planet.

Near the Earth's surface an apple is at $r \approx R_E = 6.37 \times 10^6\ \text{m}$ from Earth's centre, giving a field strength $g = GM_E/R_E^2 \approx 9.8\ \text{m s}^{-2}$. The apple's initial horizontal velocity is zero, so this acceleration simply pulls it straight down to the ground - what everyone calls "falling".

The Moon experiences exactly the same law, but at $r = 3.84 \times 10^8\ \text{m}$, roughly 60 times further from Earth's centre. Because the force falls as $1/r^2$, the Moon's gravitational acceleration toward Earth is about $60^2 \approx 3600$ times smaller than at the surface, only around $2.7 \times 10^{-3}\ \text{m s}^{-2}$. Unlike the apple, the Moon has a large tangential (sideways) velocity, so this weak acceleration does not pull it straight down - it continuously deflects the Moon's straight-line path into a closed curve, i.e. an orbit. In effect, the Moon is "falling" toward Earth forever but also moving sideways fast enough to keep missing it.

Newton's insight was that the apple and the Moon obey the identical inverse-square law of gravitation; the enormous difference in outcome (a fall versus a stable orbit) comes entirely from the difference in distance (which sets the size of $g$) and tangential speed, not from any difference in the underlying force. This was the first demonstration that the same physical law governs motion on Earth and in the heavens.

Marker's note: the top band states the SAME equation applies to both scenarios, gives a numeric estimate of $g$ (or the force) for both the apple and the Moon, and explicitly explains why a weaker, more distant force produces an orbit rather than a fall (tangential velocity plus continuous deflection). A response that only asserts "the same law applies" without the distance/magnitude reasoning caps in the middle band.

Band-6 plan. (1) State the claim to be evaluated. (2) Use the inverse-square law to show force/field strength approaches zero asymptotically but never becomes exactly zero at any finite $r$ - so mathematically the claim is false. (3) Explain why the claim FEELS true in practice (the field becomes negligibly small very quickly because of the $r^2$ term, and other, closer masses dominate). (4) Reach a balanced judgement: technically false, but a reasonable practical approximation at large distances.

Model answer. The claim is that gravity has a definite cut-off distance beyond which it is exactly zero. Newton's Law of Universal Gravitation, $F = Gm_1m_2/r^2$, shows this is not correct: for any finite separation $r$, $F$ has some non-zero value, however small. As $r \to \infty$, $F \to 0$, but this is an asymptotic limit - the force approaches zero without ever reaching it at a finite distance. The same is true of the field strength $g = GM/r^2$: it decreases smoothly and continuously with $r$, with no sudden drop to zero at any particular distance.

However, the claim captures a real physical effect: because the field falls as $1/r^2$, doubling the distance from a planet cuts the field to a quarter, and by ten times the distance the field is only one-hundredth as strong. Practically, the field becomes so small at large distances that it is completely swamped by the gravitational field of other, closer masses (for example, once far enough from Earth, the Sun's gravity dominates the motion of a spacecraft) or is simply too weak to have any measurable effect. This gives the everyday impression of a "boundary" to a planet's gravity, even though no true boundary exists.

On balance, the claim is technically false as a statement about the physics (gravity is unbounded in range and never reaches exactly zero), but it is a reasonable practical simplification: because of the rapid inverse-square fall-off, a planet's gravitational influence becomes negligible - though never mathematically zero - within a modest multiple of its own radius.

Marker's note: the top band explicitly distinguishes "approaches zero asymptotically" from "reaches zero at a finite distance", uses the $1/r^2$ dependence to justify why the field becomes negligible quickly in practice, and delivers an explicit judgement (technically false, practically reasonable) rather than a one-sided answer. A response that just says "gravity never ends" with no acknowledgement of the practical basis for the claim caps in the middle band.